# How do you find all the real and complex roots of #x^7 + 128 = 0#?

##### 2 Answers

#x^7+128 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))#

Hence roots:

#x = -2cos((2kpi)/7)-2isin((2kpi)/7)#

where

#### Explanation:

#x^7+128 = x^7+2^7#

By De Moivre's formula we find:

#(cos((2kpi)/7) + i sin((2kpi)/7))^7 = cos(2kpi) + i sin(2kpi) = 1#

for any integer

So the

#cos((2kpi)/7) + i sin((2kpi)/7)# where#k=0,1,2,3,4,5,6#

Hence we find:

#x^7+2^7 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))#

#### Explanation:

We have

Numerically we know that

(Where

But the thing is, whenever we take the nth root of a number, we bissect the circle of

That means that we can derive that the general form of the roots are

Which we can easily check back using De Moivre's formula, since

For any of the mentioned

And since